In this article we will look into what are Yoda conditions and it's advantages w.r.t C#

Introduction

Many of us has the habit of writing the programming logic as follows

Case 1:

int

a = 5;

if (a == 6){

//do something

}

or Case 2:

bool

x = false;

if (x == true){

//do something

}

That means we are using equality comparison operator to compare the values.But, if by mistake we write like

Case 1:

int

a = 5;

if (a = 6){

//do something

}

or Case 2:

bool

x = false;

if (x = true){

//do something

}

Then the whole purpose of conditional checking will be wrong.

I do agree that for the first case, C# will give a compilation error Cannot implicitly convert type 'int' to 'bool'

while for the second case the it leaves away by giving an warning Assignment in conditional expression is always constant;did you mean to use == instead of = ?

The Yoda style

A better way to write code in that case will be in the Yoda style as below

Case 1:

int

a = 5;

if (6 == a){

//do something

}

or Case 2:

bool

x = false;

if (true == x){

//do something

}

Why?Because it reduces coding errors where we unintentionally mistyped the comparison operator == as a single =

Case 1:

int

a = 5;

if (6 = a){

//do something

}

or Case 2:

bool

x = false;

if (true = x){

//do something

}

C# will give compilation error in both the cases as

The left-hand side of an assignment must be a variable, property or indexer

In simple words,we can't assign a value to a literal

Usage of this conditions style was popular on C/C++ where we could assign value to variable when typed = instead of ==

Does it really needed in C#?

Well from all the above experiments (atleast Case 1), we have seen that in C# expressions are not converted automatically to bool and henceforth this practice does not make much sence in C# parlance

However,the convention should be to always put the constant on the left hand side of the expression so that an error can be statically detected at compile time in the event of a typo e.g. = (assignment) was coded instead of the == (equality).

How IL interpretes?

To answer this question, we have developed a simple program as under

public

bool ConstantLeft(int a)

{

return a == 1;

}

publicbool ConstantRight(int a)

{

return 1 == a;

}

When viewed in IL, we found the below for ConstantLeft() function

.method

public hidebysig instance bool ConstantLeft(int32 a) cil managed

{

// Code size 10 (0xa)

.maxstack 2

.locals init ([0]

bool CS$1$0000)

IL_0000: nop

IL_0001: ldarg.1

IL_0002: ldc.i4.1

IL_0003: ceq

IL_0005: stloc.0

IL_0006: br.s IL_0008

IL_0008: ldloc.0

IL_0009: ret

}

// end of method Class1::ConstantLeft

For ConstantRight() function, IL view is as under

.method

public hidebysig instance bool ConstantRight(int32 a) cil managed

{

// Code size 10 (0xa)

.maxstack 2

.locals init ([0]

bool CS$1$0000)

IL_0000: nop

IL_0001: ldc.i4.1

IL_0002: ldarg.1

IL_0003: ceq

IL_0005: stloc.0

IL_0006: br.s IL_0008

IL_0008: ldloc.0

IL_0009: ret

}

// end of method Class1::ConstantRight

The IL infers that, line numbers IL_0001 and IL_0002, they're just swapped but the order of the operands doesn't change the behavior of ceq at IL_0003.So apart from swapping, there was no change found.